Respuesta :
Given the following System of equations:
[tex]\begin{cases}y=x^2+20x+63 \\ y=3x-9\end{cases}[/tex]You can apply the following procedure to solve it:
1. Make the equations equal to each other:
[tex]x^{2}+20x+63=3x-9[/tex]2. Solve for the variable "x":
- Make the equation equal to zero:
[tex]\begin{gathered} x^2+20x+63-3x+9=0 \\ x^2+17x+72=0 \end{gathered}[/tex]- You need to use the Quadratic formula to find the values of "x". This is:
[tex]x=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a}[/tex]In this case:
[tex]\begin{gathered} a=1 \\ b=17 \\ c=72 \end{gathered}[/tex]Substituting values and evaluating, you get:
[tex]\begin{gathered} x=\frac{-17\pm\sqrt[]{17^2-4(1)(72)}}{2(1)} \\ \\ x=\frac{-17+\sqrt[]{1}}{2} \\ \\ x=\frac{-17-\sqrt[]{1}}{2} \\ \\ x=-8;x=-9 \end{gathered}[/tex]3. Substitute each value of "x" into the second original equation and evaluate in order to find the corresponding values of "y":
- For:
[tex]x=-8[/tex]You get:
[tex]\begin{gathered} y=3(-8)-9 \\ y=-24-9 \\ y=-33 \end{gathered}[/tex]- For:
[tex]x=-9[/tex]You get:
[tex]\begin{gathered} y=3(-9)-9 \\ y=-27-9 \\ y=-36 \end{gathered}[/tex]The points are:
[tex](-8,-33);(-9,-36)[/tex]